Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}9x+4y &= 1 \\ 6x+y &= 1\end{align*}$
Answer: We can eliminate $x$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-2$ and the bottom equation by $3$ $\begin{align*}-18x-8y &= -2\\ 18x+3y &= 3\end{align*}$ Add the top and bottom equations. $-5y = 1$ Divide both sides by $-5$ and reduce as necessary. $y = -\dfrac{1}{5}$ Substitute $-\dfrac{1}{5}$ for $y$ in the top equation. $9x+4( -\dfrac{1}{5}) = 1$ $9x-\dfrac{4}{5} = 1$ $9x = \dfrac{9}{5}$ $x = \dfrac{1}{5}$ The solution is $\enspace x = \dfrac{1}{5}, \enspace y = -\dfrac{1}{5}$.